Another Surface Area question

**apw235@nyu.edu** · April 17th 2008, 07:12 PM

we have $z=1+3x+2y^2$ and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).
So far:
$dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA= \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA$ , so the integral looks ugly...

**apw235@nyu.edu** · April 18th 2008, 05:47 AM

bump

**mr fantastic** · April 18th 2008, 06:04 AM

Originally Posted by apw235@nyu.edu

we have $z=1+3x+2y^2$ and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).
So far:
$dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA= \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA$ , so the integral looks ugly...

Have you sketched the region of integration in the xy-plane?

Your integral is $S = \int_{x = 0}^{x = 2} \int_{y = x/2}^{y = 1} \sqrt{16y^2+10} \, dy \, dx$ .

Life is much easier if you reverse the order of integration:

$S = \int_{y = 0}^{y = 1} \int_{x = 0}^{x = 2y} \sqrt{16y^2+10} \, dx \, dy$ .

This should cause little trouble.

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